1,5,9......405
this is an arithmetic sequence. I think the last number is right or maybe it's 404. don't tell me you actually counted the terms.
2006-07-20 08:45:09 · 5 answers · asked by NONAME 1 in Science & Mathematics ➔ Mathematics
answer:
the last number is 405. number of terms in the seq: 102.
first number is 1.
in the sequence, the first number +4= next digit
1-5=4 which is 4*1---------------> 2nd digit is 5
1-9=8 which is 4*2
1-13=12 which is 4*3
i.e 1+4(x)=405
hence x=101
number of digits is 101+1=102
2006-07-20 09:02:54 · answer #1 · answered by human 1 · 11⤊ 4⤋
Count the n terms then
n*4-3
The n term for 405 is (405+3)/4 = 102
2006-07-20 10:03:33 · answer #2 · answered by Anonymous · 0⤊ 0⤋
1, 5, 9......405. Only for the case of counting all +3
4, 8, 12, ..408. Divide them all by 4.
1, 2 ,3 ,....102. Now you see easily you have exactly 102 terms.
2006-07-20 10:31:38 · answer #3 · answered by Thermo 6 · 0⤊ 0⤋
well.... it starts at 1 and adds four up to 405... its the same as if you added 0 to 404 by the same margin... there are 101 terms not including the first number... 102 if you include 1 (your starting number)
2006-07-20 08:51:47 · answer #4 · answered by AresIV 4 · 0⤊ 0⤋
aritmetic progression
the first term a1=1
the difference d=9-5=5-1=4
general term an=a1+(n-1)d
405=1+(n-1)4
404/4=n-1
n=101+1=102
The number of the terms is 102
And the last term is 405
2006-07-20 09:00:42 · answer #5 · answered by iyiogrenci 6 · 0⤊ 0⤋